Question

Two copper wires having their radii in the ratio of \(3:2\) are connected in series across a battery. Find the ratio of the drift velocities of the electrons in the wires.

Hint:

For wires connected in series: \[ I_1 = I_2 \] Using \[ I = neAv_d \] we get \[ v_d \propto \frac{1}{A} \] and since \[ A=\pi r^2, \] \[ v_d \propto \frac{1}{r^2} \] Therefore, drift velocity is inversely proportional to the square of the radius of the wire.

Answer 1

Concept: The drift velocity of free electrons in a conductor is related to the electric current by the relation \[ I = neAv_d \] where
• \(I\) = current flowing through the conductor,
• \(n\) = number density of free electrons,
• \(e\) = electronic charge,
• \(A\) = area of cross-section of the conductor,
• \(v_d\) = drift velocity of electrons. For conductors made of the same material, the values of \(n\) and \(e\) are identical.

Step 1: Use the condition of series connection. The two copper wires are connected in series. Therefore, the same current flows through both wires. \[ I_1 = I_2 \] Using the relation \[ I = neAv_d \] for each wire, \[ neA_1v_{d1}=neA_2v_{d2} \] Cancelling the common factors \(n\) and \(e\), \[ A_1v_{d1}=A_2v_{d2} \] Hence, \[ \frac{v_{d1}}{v_{d2}} = \frac{A_2}{A_1} \]

Step 2: Express area in terms of radius. The cross-sectional area of a wire is \[ A=\pi r^2 \] Therefore, \[ \frac{v_{d1}}{v_{d2}} = \frac{\pi r_2^2}{\pi r_1^2} \] \[ \frac{v_{d1}}{v_{d2}} = \frac{r_2^2}{r_1^2} \]

Step 3: Substitute the given ratio of radii. Given \[ r_1:r_2 = 3:2 \] Therefore, \[ \frac{v_{d1}}{v_{d2}} = \frac{2^2}{3^2} \] \[ \frac{v_{d1}}{v_{d2}} = \frac{4}{9} \] Final Result: Hence, the ratio of the drift velocities of electrons in the two wires is \[ \boxed{v_{d1}:v_{d2}=4:9} \] Thus, the thinner wire has a larger drift velocity because the same current must pass through a smaller cross-sectional area.